Chapter 16 Text

16.6 Weak Acids 631
This expression leads to a quadratic equation in x, which we can solve by using
an equation-solving calculator or by using the quadratic formula; We can also
simplify the problem, however, by noting that the value of Ka is quite small. As
a result, we anticipate that the equilibrium will lie far to the left and that x will be
very small compared to the initial concentration of acetic acid. Thus, we will
assume that x is negligible compared to 0.30, so that 0.30 - x is essentially equal
to 0.30.
0.30 -x** 0.30
As we will see, we can (and should!) check the validity of this assumption
when we finish the problem. By using this assumption, Equation 16.30 now
becomes
Solving for x, we have
0.30
= 1.8 X 10~5
.2 _ = (0.30)(1.8 x = 5.4 X 10"6
x = V5.4 X 10~6 = 2.3 X 10~3
[H+] = x = 2.3 X 10~3 M
PH = -log(Z3 X 10~3) = 2.64
We should now go back and check the validity of our simplifying assumption
that 0.30 - x — 0.30. The value of x we determined is so small that, for this
number of significant figures, the assumption is entirely valid. We are thus satisfied
that the assumption was a reasonable one to make. Because x represents
the moles per liter of acetic acid that ionize, we see that, in this particular case,
less than 1% of the acetic acid molecules ionize:
Percent ionization of HC2H3O2 = n x 100% = 0.77%
As a general rule, if the quantity x is more than about 5% of the initial value, it
is better to use the quadratic formula. You should always check the validity of
any simplifying assumptions after you have finished solving a [problem.
Finally, we can compare the pH value of this weak acid tp a solution of a
strong acid of the same concentration. The pH of the 0.30 M solution of acetic acid
is 2.64. By comparison, the pH of a 0.30 M solution of a strong acid such as HC1
is -log(O.SO) — 0.52. As expected, the pH of a solution of a weak acid is higher
than that of a solution of a strong acid of the same molarity.
When in doubt, work the problem
assuming ([HA]0 - x) ~ [HA]0. After
solving for x, compare x to [HA]0. If
x > 5% of [HA]0, rework the
problem using the quadratic
formula.
SAMPLEEXEROSE16.il
Calculate the pH of a 0.20 M solution of HCN. (Refer to Table 16.2 or Appendix D for
the value of Ka.)
Solution
Analyze and Plan: We are given the molarity of a weak acid and are asked for the pH.
From Table 16.2, Ka for HCN is 4.9 x 10~10. We proceed as in the example just worked
in the text, writing the chemical equation and constructing a table of initial and equilibrium
concentrations in which the equilibrium concentration of H+ is our unknown.
Solve: Writing both the chemical equation for the ionization reaction that forms H" (ay)
and the equilibrium-constant (Ka) expression for the reaction:
[H+1[CN~
[HCN]
- 4.9 X 10~10